Question

If a laxative manufacturer has 3.6x10^5 grams of Mg, how many grams of water must be used to have all of the Mg react to form Mg(OH)2?

A) 3.6x10^5 grams of water
B) 1.8x10^5 grams of water
C) 7.2x10^5 grams of water
D) 3.6x10^5/2 grams of water

Answer 1

To form Mg(OH)2, 1.3x10^7 grams of water must be used with 3.6x10^5 grams of Mg.

Step-by-step explanation:

Magnesium hydroxide, Mg(OH)2, is the active ingredient in milk of magnesia. The molar mass of Mg(OH)2 is 58.3 g/mol. To determine the grams of water needed to react with the given amount of Mg, we need to use the stoichiometry of the reaction.

The balanced equation for the reaction is: Mg + 2H2O -> Mg(OH)2

From the equation, we can see that one mole of Mg reacts with 2 moles of water to form one mole of Mg(OH)2. Therefore, to determine the grams of water needed, we can use the molar mass of water (18 g/mol):

(3.6x10^5 g Mg) x (2 mol water / 1 mol Mg) x (18 g / mol water) = 1.3x10^7 g of water